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CIE 9709 2026 June Paper 43 Q3

A Level / CIE / M1

CIE 9709 2026 June Paper 43 Paper · Question 3

题目

Problem

A block of mass m kgm\,\mathrm{kg} is pulled up a rough plane inclined at an angle of 30∘30^\circ to the horizontal by a constant force of magnitude 20 N20\,\mathrm{N}. This force acts parallel to a line of greatest slope of the plane. The acceleration of the block is 2 m s−22\,\mathrm{m\,s^{-2}}. The coefficient of friction between the block and the plane is 0.40.4.

Find the value of mm.

(4)
题目中文翻译

一个质量为 m kgm\,\mathrm{kg} 的物块,在大小为 20 N20\,\mathrm{N} 的恒力作用下沿粗糙斜面向上运动,斜面与水平面的夹角为 30∘30^\circ。该力平行于斜面的一条最大坡度线。物块的加速度为 2 m s−22\,\mathrm{m\,s^{-2}}。物块与斜面之间的摩擦系数为 0.40.4。

求 mm 的值。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时,取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

解法一

思路

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上行时重力沿坡分量与摩擦均向下;外力沿坡不影响正压力。

答题过程

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The normal contact force is

R=mgcos⁡30∘=53m.R=mg\cos30^\circ=5\sqrt3m.

Friction acts down the plane, with magnitude

F=0.4R=23m.F=0.4R=2\sqrt3m.

Resolving up the plane,

20−23m−m(10)sin⁡30∘=2m.20-2\sqrt3m-m(10)\sin30^\circ=2m.

Hence

20= (7+23)m,m= 207+23= 1.911296424…≈1.91.\begin{align*} 20=&\,(7+2\sqrt3)m,\\[4mm] m=&\,\frac{20}{7+2\sqrt3}\\[4mm] =&\,1.911296424\ldots\approx\boxed{1.91}. \end{align*}